A body takes 10min to cool from `60^(@)C " to" 50^(@0C`. If the temperature of surrounding is `25^(@)C`, then temperature of body after next 10 min will be

Options

A. `45^(@)C`
B. `42.85^(@)C`
C. `30^(@)C`
D. `32.50^(@)C`

Explanation

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial temperature of the body, \( T_1 = 60^\circ C \)
- Final temperature after 10 minutes, \( T_2 = 50^\circ C \)
- Surrounding temperature, \( T_0 = 25^\circ C \)
- Time interval, \( t = 10 \) minutes

2. **Apply Newton's Law of Cooling**:
The formula is given by:
\[ \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_0 \right) \]

3. **Substitute the Known Values into the Formula**:
\[ \frac{60 - 50}{10} = k \left( \frac{60 + 50}{2} - 25 \right) \]
Simplifying the left side:
\[ \frac{10}{10} = k \left( \frac{110}{2} - 25 \right) \]
\[ 1 = k \left( 55 - 25 \right) \]
\[ 1 = k \cdot 30 \]

4. **Solve for \( k \)**:
\[ k = \frac{1}{30} \]

5. **Calculate the Temperature After the Next 10 Minutes**:
Now we need to find the temperature \( T \) after another 10 minutes (from \( T_2 = 50^\circ C \) to \( T \)).
Using the same formula:
\[ \frac{T_2 - T}{10} = k \left( \frac{T_2 + T}{2} - T_0 \right) \]
Substituting the known values:
\[ \frac{50 - T}{10} = \frac{1}{30} \left( \frac{50 + T}{2} - 25 \right) \]

6. **Multiply through by 30 to eliminate the fraction**:
\[ 3(50 - T) = \frac{50 + T}{2} - 25 \]
Expanding the left side:
\[ 150 - 3T = \frac{50 + T}{2} - 25 \]

7. **Multiply through by 2 to eliminate the fraction**:
\[ 300 - 6T = 50 + T - 50 \]
Simplifying:
\[ 300 - 6T = T \]
\[ 300 = 7T \]

8. **Solve for \( T \)**:
\[ T = \frac{300}{7} \approx 42.86^\circ C \]

### Final Answer:
The temperature of the body after the next 10 minutes will be approximately \( 42.86^\circ C \).